(a) Consolidated drained tests on clay specimens were performed and the following results were obtained : Specimen No. |…

(a) Consolidated drained triaxial tests on clay (15 marks)

The initial area is A₀ = π/4 × 38² = 1134.1 mm² and the initial length L₀ = 76 mm, so V₀ = 1134.1 × 76 = 86.19 cc. In a drained test the volume change is decreasing, so the corrected area at failure is

A = A₀ (1 − ΔV/V₀)/(1 − ΔL/L₀)

• Test 1: A = 1134.1 × (1 − 7.0/86.19)/(1 − 11.0/76) = 1134.1 × 0.9188/0.8553 = 1218.3 mm². Deviator stress σ_d = 410/1218.3 = 0.3365 N/mm² = 336.5 kPa, so σ₁ = 200 + 336.5 = 536.5 kPa.
• Test 2: A = 1134.1 × (1 − 8.5/86.19)/(1 − 12.3/76) = 1134.1 × 0.9014/0.8382 = 1219.7 mm². σ_d = 850/1219.7 = 696.9 kPa, so σ₁ = 400 + 696.9 = 1096.9 kPa.
• Test 3: A = 1134.1 × (1 − 10.0/86.19)/(1 − 14.6/76) = 1134.1 × 0.8840/0.8079 = 1240.9 mm². σ_d = 1275/1240.9 = 1027.5 kPa, so σ₁ = 600 + 1027.5 = 1627.5 kPa.

The tests are drained, so the pore pressure is zero at failure: effective stresses equal total stresses (σ′₃ = σ₃ and σ′₁ = σ₁), and the total-stress and effective-stress parameters are identical.

Shear parameters. Plotting the Mohr circles (σ₃, σ₁) gives a straight envelope through the origin. On the p–q plot, p = (σ₁ + σ₃)/2 = 368.3, 748.5 and 1113.7 kPa and q = (σ₁ − σ₃)/2 = 168.3, 348.5 and 513.7 kPa. The best-fit line q = a + p tan α has tan α = sin φ′ = 0.463, so φ′ = 27.6°, with an intercept of about −1 kPa, so c′ ≈ 0. Each test alone gives φ′ = sin⁻¹[(σ₁ − σ₃)/(σ₁ + σ₃)] = 27.2°, 27.7° and 27.5°. Thus

c′ = c_d = 0, φ′ = φ_d ≈ 27.6° (total-stress values c = 0, φ = 27.6°, since u = 0).

Comment and application. The zero cohesion intercept shows a normally consolidated clay, with strength coming entirely from friction. The three φ values agree closely, so the tests are consistent and the envelope is linear over 200–600 kPa. These drained parameters govern long-term stability, when excess pore pressure has dissipated: cuts and natural slopes in clay, embankments after consolidation, and the long-term pressure on retaining walls and foundations. They do not apply to end-of-construction (undrained) conditions, for which c_u or the CU test with pore pressure measurement is used. The CD test is slow, taking days for a clay, because the shearing rate must allow full drainage.

(b) Vibroflotation and suitability number of the backfill (15 marks)

Vibroflotation. Vibroflotation is a ground-improvement method for densifying loose, clean, cohesionless soil (sand and gravel). The vibroflot is a cylindrical probe about 2 m long and 0.3–0.4 m in diameter, with an eccentric weight that gives horizontal vibrations, and water jets at the tip. It is lowered under its own weight and the jets to the required depth, up to about 15–20 m. It is then withdrawn in 0.3–0.5 m steps, holding at each step while coarse backfill is fed from the surface into the crater it makes. The vibration rearranges the grains into a denser state and the backfill fills the hole, so the ground is densified in a radius of about 1.5–2.5 m from each probe. Probes are placed on a triangular or square grid. The method raises relative density, bearing capacity, and resistance to settlement and liquefaction. It is unsuitable for soils with more than about 15–20% fines. For soft clays, the same equipment is used with stone backfill as vibro-replacement, forming stone columns.

Suitability number. The total mass is 9.4 + 54.0 + 78.2 + 83.1 + 86.2 + 75.3 + 13.8 = 400 g. The cumulative percentage finer is

• 4.75 mm: 97.65%
• 2.36 mm: 84.15%
• 1.18 mm: 64.60%
• 0.600 mm: 43.83%
• 0.300 mm: 22.28%
• 0.150 mm: 3.45%
• 0.075 mm: 0% (nothing finer than 75 μm remains after the last sieve)

Interpolating on the semilog grading curve:

• D₁₀ lies between 0.150 mm (3.45%) and 0.300 mm (22.28%): D₁₀ ≈ 0.191 mm.
• D₂₀ lies in the same interval: D₂₀ ≈ 0.276 mm.
• D₅₀ lies between 0.600 mm and 1.18 mm: D₅₀ ≈ 0.734 mm.

The suitability number is

SN = 1.7 √(3/D₅₀² + 1/D₂₀² + 1/D₁₀²) = 1.7 √(3/0.538 + 1/0.0761 + 1/0.0364)

SN = 1.7 √(5.57 + 13.14 + 27.4) = 1.7 × 6.79 ≈ 11.5

Rating. The Brown rating scale is: 0–10 excellent, 10–20 good, 20–30 fair, 30–50 poor, above 50 unsuitable. An SN of about 11.5 lies in the 10–20 band, so the backfill is rated good, close to the excellent boundary.

(c)(i) Francis turbine: inlet triangle, discharge and power (15 marks)

Inlet blade speed u₁ = πD₁N/60 = π × 1.20 × 300/60 = 18.85 m/s.

Inlet velocity triangle. The blade angle at inlet is 90°, so the relative velocity is radial and the whirl velocity equals the blade speed: V_w1 = u₁ = 18.85 m/s. The guide vane angle is α₁ = 30°, so the flow velocity is V_f1 = V_w1 tan 30° = 18.85 × 0.5774 = 10.88 m/s. The absolute velocity is V₁ = V_w1/cos 30° = 21.77 m/s.

Discharge. With the vane thickness coefficient k = 0.95: Q = k π D₁ B₁ V_f1 = 0.95 × π × 1.20 × 0.28 × 10.88 = 10.91 m³/s.

Work done. The discharge at the outlet is radial, so V_w2 = 0 and the work done on the runner per unit weight of water is V_w1 u₁/g = 18.85²/9.81 = 36.2 m. The power developed by the runner is

P_runner = ρQ V_w1 u₁ = 1000 × 10.91 × 18.85 × 18.85 = 3.88 × 10⁶ W = 3.88 MW.

Power produced. With the overall efficiency of 90%: P = 0.90 × 3.88 = 3.49 MW ≈ 3.5 MW.

(c)(ii) Curves of constant stream function (5 marks)

Significance. For two-dimensional incompressible flow, the stream function ψ(x, y) is defined by u = ∂ψ/∂y and v = −∂ψ/∂x. A curve ψ = constant is a streamline: the velocity is tangent to it and no fluid crosses it. The difference in ψ between two streamlines equals the volume flow per unit depth between them.

Why it is useful. The stream function satisfies continuity automatically, so it reduces the two velocity unknowns to one. For irrotational flow ψ satisfies Laplace's equation, is orthogonal to the equipotential lines of the velocity potential (which gives the flow net), and obeys linear superposition, so uniform flow, sources, sinks, vortices and doublets can be combined to model flow past bodies. A solid boundary is a streamline, which makes boundary conditions simple. Closely spaced streamlines show high velocity, so a plot of ψ gives a picture of the flow field.